anonymous
  • anonymous
The second derivative of (4-x^2)*exp(-1.16x^2)
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
can anyone help me with this?? =]
anonymous
  • anonymous
you can use the product rule right?
anonymous
  • anonymous
h(x)=f(x)g(x) h(x)=f(x)g(x)=f(x)g(x)+f(x)g(x)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
h'(x) = [f(x)g(x)]' = f'(x)g(x)+f(x)g'(x)
anonymous
  • anonymous
yes but it doesn't work. my online program does not accept it. (4-x^2)(5.3824*^2*exp(-1.16x^2)-(-2.32xexp(-1.16x^2)(-2x)-(exp(-1.16x^2)(-2)-(-2x)(-2.32xexp(-1.16x^2)
anonymous
  • anonymous
is that what you got?
anonymous
  • anonymous
I didn't work it out. I was just pointing out that you can use the product rule for this kind of problem.
anonymous
  • anonymous
i did. do you think you can check my work?
anonymous
  • anonymous
but your solution seems correct. see if you can simplify it further. Maybe your online tool will accept it then
anonymous
  • anonymous
I think I am getting a slightly different answer from you. check your calculations.
anonymous
  • anonymous
what is slightly different about it? I just double checked!
anonymous
  • anonymous
did you simplify it. wait let me post my solution
anonymous
  • anonymous
you can set x^2 = y and factor the second term.
anonymous
  • anonymous
as a quadratic equation of course.

Looking for something else?

Not the answer you are looking for? Search for more explanations.